<Question>
Evaluate $\log_264$.
</Question>

<think>
<Plan_1>
Okay, let's see. I need to evaluate log base 2 of 64. Hmm, logarithms can sometimes be tricky, but maybe if I break it down step by step, it will make more sense. So, log base 2 of 64 is asking the question: "2 raised to what power gives me 64"? Right?

First, I remember that logarithms and exponents are related. So, if I have log₂(64) = x, then that means 2^x = 64.

Now, I need to figure out what exponent x makes 2^x equal to 64. Let me start by recalling the powers of 2.

<prompt_1.1> Recall the powers of 2 to find the exponent that results in 64. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, and 2^6 is 64. Oh, there it is! So, 2^6 equals 64.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, let me double-check that.
2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, 2^6 = 64. Yep, that's correct. So x must be 6.

Therefore, log base 2 of 64 is 6.

But maybe there's another way to approach this in case I can't recall the powers of 2 off the top of my head. Let's think about prime factorization.

<prompt_2.1> Use prime factorization to find log base 2 of 64. </prompt_2.1>

Alternatively, maybe using the change of base formula. The logarithm log_b(a) can be calculated as log_c(a)/log_c(b) for any positive c ≠ 1. If I use base 10, which is common, then log₂(64) = log(64)/log(2).

<prompt_2.2> Use the change of base formula to calculate log base 2 of 64. </prompt_2.2>

Another approach: maybe expressing 64 as a power of 2. Since 64 is 8^2, but 8 is 2^3, so 64 = (2^3)^2 = 2^(3*2) = 2^6. Therefore, log₂(64) = 6.

Wait, what if I use exponents rules? If I let log₂(64) = x, then 2^x = 64. Since 64 is a power of 2, as established before, so x must be 6.

Is there a possibility I made a mistake in my reasoning? Let's check with a different base. Suppose I use natural logarithm.

<prompt_2.3> Verify the result using natural logarithms. </prompt_2.3>

Alternatively, maybe using binary logarithm since base 2 is related to binary. The binary logarithm of 64 is the number of bits needed to represent 64 in binary. Since 64 in binary is 1000000, which is 1 followed by six 0s, so that's 7 bits. Wait, but the value is 2^6 = 64, so the exponent is 6, but the binary representation is 7 bits.
Hmm, maybe that's a red herring. But the logarithm value is still 6, because log₂(64) is 6, regardless of the bit count. Because 2^6 is 64, so the exponent is 6.

Wait, another way to think about it: If I have 2^x = 64, then taking the logarithm base 2 of both sides gives x = log₂(64). But that's just restating the original problem. So, to solve for x, we need to evaluate log₂(64), which is 6 as we've already determined.

Let me consider if there's any alternative interpretation of the question. Is log₂64 ambiguous? Well, the notation log₂64 without parentheses is standard for log base 2 of 64. If it were log base 2 of 6 followed by 4, it would be written differently, but as written, it's log₂(64).

<prompt_2.4> Consider any potential ambiguities in the notation and conclude the solution. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
64 is 64.0, right? So, dividing by 2 each time: 64 divided by 2 is 32, then 16, then 8, then 4, then 2, then 1. So that's 6 divisions by 2, which means 2^6 = 64. So again, x is 6.
</execution_2.1>

<execution_2.2>
Calculating that, let's see. If I use a calculator, log(64) is approximately 1.80618 (assuming natural logarithm, but actually, in programming terms, log could be base e or base 10 depending on context, but in mathematics, log without a base is often base 10, while ln is natural logarithm). Wait, actually, in higher mathematics, log often refers to base e, but in many contexts, especially in engineering and some sciences, log can be base 10. However, when changing the base, the formula works for any base. So, to avoid confusion, maybe I should use ln for natural logarithm. Let me clarify: log₂(64) = ln(64)/ln(2). Let's compute that.

First, ln(64). Since 64 is 2^6, ln(64) = ln(2^6) = 6*ln(2). Therefore, ln(64)/ln(2) = (6*ln(2))/ln(2) = 6. So that also gives 6. So, regardless of the base I choose for the change of base formula, as long as I use the same base for both logarithms, it simplifies correctly. That's a good check.
</execution_2.2>

<execution_2.3>
Then log₂(64) = ln(64)/ln(2). I know that ln(64) is ln(2^6) = 6*ln(2). Therefore, ln(64)/ln(2) = 6*ln(2)/ln(2) = 6. So that also gives 6.
</execution_2.3>

<execution_2.4>
So, all methods point to the answer being 6. I don't think I made any mistakes here. All the different approaches—recalling powers of 2, using prime factorization, change of base formula, and even binary representation—lead to the same conclusion. Therefore, I'm confident that log base 2 of 64 is 6.
</execution_2.4>
</Execution_2>

<Final_answer>
Therefore, $\log_264 = \boxed{6}$.
</Final_answer>
</think>